Question 1118350
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In this systems of equations, how can I figure out what the 2nd equation's minimum/least dollar value is?

Purchases of $6 or $4 each month. Is there a better way to solve this?
Thanks in advance.

m+s=12

6m+4s=?
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The second equation is incomplete; it is only an expression, but not an equation.  What are the variables, m and s, supposed to be?  quantity counts for $6 and $4 ?  If they are, then  {{{m>=0}}} and {{{s>=0}}} and each is only whole-numbers or 0.


If vertical axis counts m and horizontal axis counts s then graph for first equation is 
{{{graph(300,300,-1,12,-1,12,12-x)}}}.
m for vertical axis and s for horizontal axis;
counts y and counts x for m and for x, respectively.


{{{6m+4s}}} does not establish any constrain unless it is equated to some value.  


Are you trying to make something like   {{{f(m)=6m+4s}}} with {{{m+s=12}}} as a restriction?
{{{f(m)=6m+4(12-m)}}}
{{{f(m)=6m+48-4m}}}
{{{f(m)=2m+48}}}
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{{{df/dm=2}}}, but then this is a constant, and {{{df/dm<>0}}}.
No minimum for f and no maximum for f.


The SMALLEST value that m can be is  0, and this will make  {{{highlight_green(f(m)=f(0)=2*0+48=highlight(48))}}}.